Question

If ${\tan }^{-1}\left(\alpha +i\beta \right)=x+iy$ then $x=$

• A. $\left(\frac{1}{2}\right){\tan }^{-1}\frac{\left(2\alpha \right)}{\left(1-{\alpha }^2-{\beta }^2\right)}$
• B. $\left(\frac{1}{2}\right){\tan }^{-1}\frac{\left(2\alpha \right)}{\left(1+{\alpha }^2+{\beta }^2\right)}$
• C. ${\tan }^{-1}\frac{\left(2\alpha \right)}{\left(1-{\alpha }^2-{\beta }^2\right)}$
• D. None of these

Answer 1

The correct option is A

$\left(\frac{1}{2}\right){\tan }^{-1}\frac{\left(2\alpha \right)}{\left(1-{\alpha }^2-{\beta }^2\right)}$

Explanation for the correct option:

Given information:

${\tan }^{-1}\left(\alpha +i\beta \right)=x+iy$ …(1)

Write the conjugate and solve

${\tan }^{-1}\left(\alpha -i\beta \right)=x-iy$ …(2)

Add equations (1) and (2)

$\Rightarrow x+iy+x-iy={\tan }^{-1}\left(\alpha +i\beta \right)+{\tan }^{-1}\left(\alpha -i\beta \right)$

$\begin{array}{l}\Rightarrow 2x={\tan }^{-1}\left(\alpha +i\beta \right)+{\tan }^{-1}\left(\alpha -i\beta \right)\\ \Rightarrow x=\left(\frac{1}{2}\right){\tan }^{-1}\frac{\left(\alpha +i\beta +\alpha -i\beta \right)}{\left(1-\left(\alpha +i\beta \right)\left(\alpha -i\beta \right)\right)}\\ \Rightarrow x=\left(\frac{1}{2}\right){\tan }^{-1}\frac{\left(2\alpha \right)}{\left(1-{\alpha }^2-{\beta }^2\right)}\end{array}$

Hence option (A) is correct.